A soldier jumps out from an aeroplane with a parachute. After dropping through a distance of 19.6 m, he opens the parachute and decelerates at the rate of 1 ms -2 . If he reaches the ground with a speed of 4.6 ms -1 , how long was he in air?
Text Solution
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Let t 1 be the time before opening of parachute.
Using h = ut + 1/2 gt 2 , we get
19.6 = 0 + 1/2 × 9.8 × t 1 2
or t 1 2 = 19.6/4.9 = 4 or t 1 = 2 s
Taking v 1 as velocity attained after falling through 19.6 m and using v 2 – u 2 = 2gh, we have
v 1 2 - 0 = 2 × 9.8 × 19.6
Again taking t 2 as time taken after opening of parachute and using v = u + at, we get
4.6 = 19.6 – 1 × t 2
or t 2 = 19.6 – 4.6n = 15 s
∴ ∴ total time = t 1 + t 2 = 2 + 15 = 17s.
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